Friction solved problems
Calculating friction and pulley problems notes
▲
According to the second law of motion we have
F = m a
In the gravitational system the force acting on a mass is
W = m g
[
kgm m / s
^{2}
]
A force of 1 Newton is equal to
N = 1
[
kgm m / s
^{2}
]
Then if the
weight
is given it equals
W
[
kgf
]
= 9.8 N
There are two friction coefficients the static μ
_{s}
and the the dynamic or kinetic μ
_{k}
The static friction coefficient value is greater then the kinetic coefficient value ( μ
_{s}
> μ
_{k}
), it means that we have to employ greater force in order to start the movement of the mass, but once the mass is moving we have to employ less force to keep it moving.
In many problems we assume that both coefficients are the same, if not we have to relate to the proper coefficient e.g. if the bodies are moving we have to take the kinetic friction coefficient.
In many questions we have to decide the direction of the movment, it is important because the direction of the friction force is always opposite to the movment direction.
If we look at the figure at left the direction of the movment is to the right as the direction of a
_{1}
the equilibrium equation will be:
ΣF
_{x}
= F − m g sin θ − f = m a
and the acceleration is:
If the movment is downward as in the second figure at left then the equilibrium equation will be:
ΣF
_{x}
= m g sin θ − F − f = m a
and the acceleration is:
We can clearly see the difference of the signs and the values of the accelerations in both direction of the chose directions.
The conclusion frome this calculation is that in friction problems a negative acceleration result doesn't mean that the acceleration is in the opposite direction but rather we have to change the direction of the acceleration and solve again, if this time we got negative value then the mass will not move due to the friction force, for more information see
example 7
,
example 38
.
If mass M is moving a distance of x meter then this distance is divided by each cable that holds mass m and mass m is moving a distance of x/2 meter.
The
distance traveled
is given by:
x = v
_{0}
t + a t
^{2}
/ 2
At rest v
_{0}
= 0 and the acceleration is:
a = 2 x / t
^{2}
time t of both masses M and m is the same so we can see that the acceleration is related to the distance, if the distance is half also the acceleration will be half so:
a
_{1}
= 2 a
_{2}
For system rest we need that the acceleration will be 0. From the equation
v
_{t}
= v
_{0}
+ a t
we get the result
v
_{t}
= v
_{0}
Friction example - 1
▲
A body whose mass is m is resting on an inclined surface with an angle of θ and a friction coefficient of μ, find the acceleration of the mass m as a function of μ and angle θ.
From equilibrium in the x and y direction we get:
ΣF
_{x}
= m g sinθ − N μ = m a
(1)
ΣF
_{y}
= N − m g cosθ = 0
(2)
From eq.
(2)
N = m g cosθ
Substitute N into eq.
(1)
m g sinθ − m g μ cosθ = m a
And the acceleration is:
a = g( sin θ − μ cos θ)
When the system is at rest a = 0 then:
We can see that the friction coefficient for rest condition must be equal or greater then the tangent of the slope μ > tan θ.
Friction example - 2
▲
A body whose mass is m is resting on a horizontal surface whose friction coefficient is μ, a force F is implied on the mass at an angle of θ degree. Find the acceleration of mass m and the maximum force that the mass m will still be at rest.
From the forces diagram we can derive the equilibrium equations:
ΣF
_{x}
= F cos θ − Nμ = ma
(1)
ΣF
_{y}
= N + F sin θ − mg = 0
(2)
From eq.
(2)
N = mg − F sin θ
Substitute N into eq.
(1)
F cos θ − mgμ + Fμ sin θ = ma
Solving for a we get:
(3)
For the maximum force F that can be applied so that the mass m will still be at rest we have:
f
_{max}
= N μ
_{s}
= (m g − F sinθ) μ
_{s}
(4)
(5)
A force of 30 N is applied to a mass of 3 kg for 5 sec at an angle of 30 degree, if the static and kinetic frictions are 0.5 and 0.2 respectively. Find the acceleration and the time needed for mass m to stop after force F is eliminated.
First check if the mass will move by the condition: f
_{max}
<
F cosθ.
Because 7.2
<
25.98 the mass will move and the acceleration is according to equation (3).
The velocity of the mass after 5 sec is found from the
acceleration aquation
:
v
_{t}
= v
_{0}
+ a t = 0 + 7.7 * 5 = 38.5 m/s
The new free body diagram after eliminating force F is at left.
According to the second law of motion F = m a (F now is only the friction force) and f = − m a (the minus sign is because of the opposite directions of a and f).
a = − f / m = − m g μ
_{k}
/ m = −g μ
_{k}
= − 9.8 * 0.2 = − 1.96 m/s
^{2}
And the time untill mass m will stop is according to the equation: v
_{t}
= v
_{0}
+ a t where v
_{t}
= 0 and v
_{0}
= 38.5 m/s
t = − v
_{0}
/ a = − 38.5 / − 1.96 = 19.64 sec
Friction example - 3
▲
A body whose mass is m is resting on a slope surface whose friction coefficient is μ and with an angle of θ, a force F is implied on the mass at an angle of α degree. Finf the acceleration of mass m and the maximum force that the mass m will still be in rest.
We will choose a positive acceleration upward and define δ = α + θ.
From the forces diagram we can derive the equilibrium equations:
ΣF
_{x}
= F cos δ − mg sin θ − Nμ = ma
(1)
ΣF
_{y}
= N − mg cos θ − F sinδ = 0
(2)
From eq. (2) we have:
N = mg cos θ + F sin(α + θ)
If the result of a is negative then the acceleration is in the opposite direction.
The maximum force for rest condition (when a = 0) is:
Check the case that the direction of the force F is upward as shown below.
The friction force is:
f = Nμ = (F sinδ − m g cosθ) μ
(angle δ = α − θ)
From the forces diagram we can derive the equilibrium equations in the slope direction, we assume that the motion is upward if the result of a is negative then we have to change the directiom of the motion.
ΣF
_{x}
= F cosδ − f − mg sin θ = m a
(3)
and the acceleration is:
If a = 0 then:
Notice that we got the same answer as before but with some variations in the signs.
Check the case that the direction of the force F is downward as shown below.
The friction force is:
f = Nμ = ( m g cosθ + F sinδ) μ
(angle δ = α − θ)
From the forces diagram we can derive the equilibrium equations in the slope direction:
ΣF
_{x}
= F cosδ − f + mg sin θ = m a
(4)
and the acceleration is:
(5)
Find the force F that should be applied in order to bring the weight of 100 kg to a velocity of 30 m/s in 20 m neglect the friction.
From the acceleration equations we can find the acceleration: v
_{t}
^{2}
= v
_{0}
^{2}
+ 2 a d
After substituting v
_{0}
= 0 we get a = v
_{t}
^{2}
/ (2 d) = 900 / (2 * 20) = 22.5 m / s
^{2}
From eq. (5) we get:
Friction of two masses example - 4
▲
Two masses are located on each other, mass m is tied to the wall by a rope, a force F is exerted on mass M , find the ecceleration of mass M if the friction coefficients between the two masses is μ
_{1}
and μ
_{2}
between mass M and the lower surface.
Friction force f
_{1}
is:
f
_{1}
= mgμ
_{1}
Friction force f
_{2}
is:
f
_{2}
= (M + m)gμ
_{2}
Mass M is effected by both frictions f
_{1}
and f
_{2}
and is:
f
_{M}
= f
_{1}
+ f
_{2}
= mgμ
_{1}
+ (M + m)gμ
_{2}
From the forces diagram we can derive the equilibrium equations in the x direction on both masses:
On mass m:
ΣF
_{x}
= T
_{1}
− mgμ
_{1}
= 0
(1)
On mass M:
ΣF
_{x}
= F − f
_{1}
− f
_{2}
= Ma
(2)
Maximum force F that can be applied to mass M without sliding is:
F <= mgμ
_{1}
+ (M + m)gμ
_{2}
From eq (1) T
_{1}
can be found:
T
_{1}
= mgμ
_{1}
Friction of two masses example - 5
▲
Mass m
_{1}
is resting on mass m
_{2}
and has a friction coefficients of μ
_{1S}
μ
_{1k}
the friction coefficients of m
_{2}
and the lower surface are μ
_{2S}
μ
_{2k}
. Find the biggest force F that can be applied to m
_{2}
without causing mass m
_{1}
to slip.
m
_{1}
kg
m
_{2}
kg
μ
_{1S}
μ
_{1k}
μ
_{2S}
μ
_{2k}
F
N
a
_{1}
m/s
_{2}
a
_{2}
m/s
_{2}
f
_{1}
N
f
_{2}
N
Force needed to start moving the masses is:
F = (m
_{1}
+ m
_{2}
) g μ
_{2S}
Mass m
_{1}
is attached to m
_{2}
only by the maximum friction force:
f
_{1}
= m
_{1}
g μ
_{1S}
this force is causing mass m
_{1}
to accelerate to its maximum value before slipping:
ΣF
_{x}
on m
_{1}
m
_{1}
g μ
_{S1}
= m
_{1}
a
_{1}
⟶
a
_{1}
= g μ
_{1S}
(1)
From the free body diagram of mass m
_{2}
we have
ΣF
_{x}
on m
_{2}
F − f
_{1}
− f
_{2}
= m
_{2}
a
_{2}
F − m
_{1}
g μ
_{1S}
− (m
_{1}
+ m
_{2}
) g μ
_{2k}
= m
_{2}
a
_{2}
(2)
Solving for the value of a
_{2}
from eq. (2) we get:
(3)
In order to prevent from mass m
_{1}
to slip both accelerations a
_{1}
and a
_{2}
should be the same a
_{1}
= a
_{2}
and the maximum force F is:
F = (m
_{1}
+ m
_{2}
)(a
_{1}
+ g μ
_{2k}
)
(4)
If the applied force F is greater then the value of F in eq. (4) then the mass m
_{1}
will slide on top of mass m
_{2}
and the friction force f
_{1}
will be the dynamic force which is less then the static force because μ
_{s}
> μ
_{d}
so the forces on mass m
_{2}
in this case is:
F − m
_{1}
gμ
_{1k}
− (m
_{1}
+ m
_{2}
)gμ
_{2k}
= m
_{2}
a
_{2}
Q.
Mass of 3 kg is mounted on top of a 7 kg mass, the friction coefficients of the upper mass are μ
_{1S}
= 0.8 and μ
_{1k}
= 0.1 and the lower mass friction coeficients are μ
_{2S}
= 0.2 and μ
_{2k}
= 0.1 . Find the maximun force F in case that mass m
_{1}
will not slide.
S.
First find the maximum acceleration acting on m
_{1}
with no sliping according to eq. (1):
a
_{1}
= g μ
_{1S}
= 9.8 * 0.8 = 7.84 m/s
^{2}
This is also the maximum acceleration of mass m
_{2}
according to eq. (4) with the assumption that a
_{1}
= a
_{2}
F = (m
_{1}
+ m
_{2}
)(a
_{1}
+ g μ
_{2k}
) = 88.2 N
Friction of two masses example - 6
▲
Find the acceleration of the system of masses neglecting the mass of the string and the inertia of the pulley.
The kinetic friction force is equal to:
f = μ
_{k}
N = μ
_{k}
M g
From the free body diagram on mass m and M and assuming that the acceleration is downward we get:
On mass M:
ΣF
_{x}
= T − M g μ
_{k}
= M a
(1)
On mass m:
ΣF
_{y}
= m g − T = m a
(2)
Eliminating T from both equations gives:
m g − M g μ
_{k}
+ M a = −m a
(3)
Mass M will accelerate when
m > μ
_{s}
M
System of the masses are at rest when
m <= μ
_{s}
M (then a = 0)
Find the friction coefficient and the tension in the rope if M = 5 kg and m = 1 kg the masses are moving at a constant speed.
Because the masses are moving at a constant speed it is necessary that a = 0. from eq. (3) we have
m − M μ
_{k}
= 0
The tension in the rope can be found by solving eq. (1) and (2) :
Friction of two masses example - 7
▲
Find the acceleration of the system of masses, the masses of the rope and the pulleys are negligible, assume first that the acceleration of M is upward and second time that the acceleration is downward.
The friction force is equal to:
f = μ N = μ M g cosθ
From the forces diagram on mass m and M and assuming that the acceleration is upward (Figure 1) we get:
On mass M:
ΣF
_{x}
= T − M g sinθ − f = M a
(1)
On mass m:
ΣF
_{y}
= m g − T = m a
(2)
(3)
Solving the case with the acceleration downward (Figure 2).
On mass M:
ΣF
_{x}
= M g sinθ − T − f = M a
(4)
On mass m:
ΣF
_{y}
= T − m g = m a
(5)
(6)
Check the case when M = 2 kg m = 1.2 kg angle θ = 30 degree and the friction coefficient is 0.15
Because both accelerations in both directions are negative motion is not possible in this conditions of masses, friction and slope angle (see note 2).
Determine the acceleration and the motion direction if M = 4 kg
m = 2 kg angle θ = 45 degree and the friction coefficient is 0.1
First check the case of upward acceleration (Figure 1).
Because a is negative and the present of friction we have to make the calculation again assumming this time that the acceleration is downward see the case of (Figure 2).
Notice that we got a different value for the acceleration, the conclution is that mass M is slidding downward.
Friction of two masses example - 7
▲
Figure 1
Figure 2
M
kg
m
kg
θ
deg
μ
a
m/s
_{2}
T
N
f
N
Find the acceleration of the system of masses, the masses of the rope and the pulleys are negligible, assume first that the acceleration of M is upward and second time that the acceleration is downward.
The friction force is equal to:
f = μ N = μ M g cosθ
From the forces diagram on mass m and M and assuming that the acceleration is upward (Figure 1) we get:
On mass M:
ΣF
_{x}
= T − M g sinθ − f = M a
(1)
On mass m:
ΣF
_{y}
= m g − T = m a
(2)
(3)
Solving the case with the acceleration downward (Figure 2).
On mass M:
ΣF
_{x}
= M g sinθ − T − f = M a
(4)
On mass m:
ΣF
_{y}
= T − m g = m a
(5)
(6)
Check the case when M = 2 kg m = 1.2 kg angle θ = 30 degree and the friction coefficient is 0.15
Because both accelerations in both directions are negative motion is not possible in this conditions of masses, friction and slope angle (see note 2).
Determine the acceleration and the motion direction if M = 4 kg
m = 2 kg angle θ = 45 degree and the friction coefficient is 0.1
First check the case of upward acceleration (Figure 1).
Because a is negative and the present of friction we have to make the calculation again assumming this time that the acceleration is downward see the case of (Figure 2).
Notice that we got a different value for the acceleration, the conclution is that mass M is slidding downward.
Note 1:
from eq. (3) and (6) and comparing the acceleration to zero we can derive the range of mass m
that the system will stay at rest.
M(sinθ − μ cosθ) < m < M(sinθ + μ cosθ)
Note 2:
to determine the direction of the motion we can solve the equilibrium equation by eliminating the friction, once we have the direction of the motion we can set the correct direction of the friction force (always opposite to the motion direction) and solve the equilibrium equations. We have to remember that if we found motion in any direction without friction, the friction force can reduce the motion acceleration or even stope the motion.
Friction of two masses example - 8
▲
Find the acceleration of the system of masses neglecting the mass of the string and the inertia of the pulley.
From the forces diagram on mass m and M and assuming that the acceleration is downward hance negative we get:
On mass M:
ΣF
_{x}
= T − Mgμ
_{1}
= −Ma
(1)
On mass m:
ΣF
_{y}
= mg sinθ + mgμ
_{2}
cosθ − T = −ma
(2)
Substitute the value of T from eq. 1 to the second eq.
System is at rest when:
Mμ
_{1}
− m(sinθ + μ
_{2}
cosθ) = 0
Two masses on inclined surfaces example - 9
▲
Find the acceleration of the system of masses neglecting the mass of the rope and the inertia of the pulley, the frictions coefficients and the slopes are given, also find the tension in the rope.
To determine the possible direction of the motion we will first solve the forces equations by neglecting the frictions, the results are the conditions:
m sinβ > M sinα
motion can be to the right
(a)
m sinβ < M sinα
motion can be to the left
(b)
m sinβ = M sinα
motion not possible
(c)
Those conditions alone are not enough to verify if the system will move, in order to make the system to slipe the masses have to overcome the friction forces, so the conditions for motions are.
m ( sinβ − μ
_{2s}
cosβ ) > M ( sinα + μ
_{1s}
cosα )
motion is to the right
(d)
M ( sinα − μ
_{1s}
cosα ) > m ( sinβ − μ
_{2s}
cosβ )
motion is to the left
(e)
From the forces diagram and assuming that the acceleration is to the right we get:
On mass M:
ΣF
_{x}
= T − M g sinα − M g cosα μ
_{1k}
= M a
(1)
On mass m:
ΣF
_{x}
= m g sinβ − T − m g cosβ μ
_{2k}
= m a
(2)
System is at rest when:
m(sinβ − μ
_{2s}
cosβ) − M(sinα + μ
_{1s}
cosα) = 0
In order to get positive tension in the rope we need that at least one of the equations will be true:
μ
_{1s}
< tan
α
or
μ
_{2s}
< tan
β
Given two masses M = 40 kg and m = 20 kg, connected with an ideal pulley, the masses are located on two surfaces whose friction coefficients are μ
_{1s}
= μ
_{1k}
= 0.15 and μ
_{2s}
= μ
_{2k}
= 0.25, the slopes has angles of α = 30 deg and β = 53 deg. Find the direction and the value of the acceleration and the tension in the rope which connects both masses.
M
kg
α
deg
m
kg
β
deg
μ
_{1s}
μ
_{1k}
μ
_{2s}
μ
_{2k}
a
m/s
^{2}
T
N
No friction move:
Each mass move:
M
m
System move:
Line
Possibilities
Description
No friction move
←
→
Checks the direction of motion in the case that
the friction is eliminated from the system.
Each mass move
M
m
Checks if motion occurs for each mass separately in the case that the rope connecting both masses is eliminated and only friction forces are present μ < tanα
* Green M is to mark that M will slide.
* Red m is to mark that m will not slide.
System move
Slide left
Slide right
No motion
Describes if the system of masses will slide to the left or to the right or it will not slide at all.
Notes:
1.
When system moves to the right or left the calculation of the tension is straightforward according to the equations found before.
2.
When the system has no motion the maximum tension can be from 0 to the minimum static friction force calculated separately for each mass (see note 3).
3.
In the case of no motion we have to analyze the conditions of the 'Each mass move'. if both masses are red then the tension can be artificially set from 0 to the minimum friction value between both masses (not the maximum value because this will cause the mass of minimum value to move and reduce the tension). If one mass is red and the other is green then the tension is the value of the green mass friction (minimum friction force).
First determine the possible motion direction according to creteria (a) and (b)
m sinβ = 20 sin53 = 16
<
M sinα = 40 sin30 = 20
The direction of the acceleration will be to the left.
Now check creterion (c) and (d) to verify is motion occures:
Motion equations for the case with acceleration to the left are:
On mass M:
ΣF
_{x}
= M g sinα − T − M g cosα μ
_{1k}
= M a
(3)
On mass m:
ΣF
_{x}
= T − m g sinβ − m g cosβ μ
_{2k}
= m a
(4)
Because we already found that the motion is to the left we have to check condition (d) only and we get 14.8 > 12.9 so the masses will slip to the left and the acceleration is:
Friction of two masses example - 10
▲
Find the acceleration of the system of masses neglecting the mass of the string and the inertia of the pulley.
From the forces diagram on mass m and M and assuming that the acceleration is to the right hance negative we get:
On mass m
_{1}
ΣF
_{x}
= T
_{1}
− m
_{1}
gμ
_{1}
= m
_{1}
a
(1)
On mass m
_{2}
ΣF
_{xy}
= T
_{2}
+ m
_{2}
g sin θ − T
_{1}
− m
_{2}
g μ
_{2}
cosθ = m
_{2}
a
(2)
On mass m
_{3}
ΣF
_{y}
= m
_{3}
g − T
_{2}
= m
_{3}
a
(3)
We get 3 equations with three unkowns T
_{1}
T
_{2}
and the acceleration a, solving for a we get:
System is at rest when a = 0:
m
_{3}
g − m
_{1}
gμ
_{1}
− m
_{2}
gμ
_{2}
cosα = 0
Friction of two masses and pulley - example 11
▲
Find the acceleration and the tension in the rope of the system of masses shown
(
M
>
m
)
, neglecting the mass of the string and the inertia of the pulley assume that the static and kinetic friction coefficients are equal to μ
_{1}
μ
_{2}
and the slope angle θ are known.
Notice that the two masses are moving in the oposite directions so the friction forces are:.
Friction forces on mass M are:
f
_{1}
= (M + m)gμ
_{1}
cosθ
Friction forces on mass m are:
f
_{2}
= mgμ
_{2}
cosθ
From the forces diagram on mass m and M and assuming that the acceleration of mass M is downward and is equal to the acceleration of mass m in the upward direction we get:
On mass m
ΣF
_{x}
= T − f
_{2}
− mg sinθ = ma
(1)
On mass M
ΣF
_{x}
= Mg sinθ − f
_{1}
− f
_{2}
− T = Ma
(2)
Subsitute the values of f
_{1}
and f
_{2}
into eq. 1 and 2 and eliminating T from both equations we get a:
Friction of two masses example - 12
▲
Find the acceleration of the system of masses neglecting the mass of the rope and the inertia of the pulley.
From the forces diagram on mass m and M and assuming that the acceleration is to the downward direction of mass M we get:
On mass M:
ΣF
_{y}
= Mg − T = Ma
(1)
On mass m:
ΣF
_{y}
= T − mg = ma
(2)
Solving for the acceleration we get:
And the tensions in the cables are:
Friction of two masses example - 13
▲
Find the acceleration of the system of masses neglecting the mass of the rope and the inertia of the pulley, the friction coefficient between mass M and the surface is μ.
From the forces diagram on mass m and M and assuming that the acceleration is to the downward direction and notice that the acceleration of mass M is twice the accelaretion of mass m because when mass M is moving a length of x mass m will travel only half this distance.
Forces on mass M:
ΣF
_{x}
= T − Mgμ = Ma
_{1}
(1)
Forces on mass m:
ΣF
_{y}
= mg − T
_{1}
= ma
_{2}
(2)
From acceleration:
a
_{1}
= 2a
_{2}
(3)
From the pulley:
T
_{1}
= 2T
(4)
We got four equations with four unkowns T T
_{1}
a
_{1}
and a
_{2}
, solving the equations we get:
The accelerations are:
The tensions are:
Friction of two masses example - 14
▲
Find the acceleration of the system of masses neglecting the mass of the rope and the inertia of the pulleys.
Forces on pulley 1:
ΣF
_{y}
= 2T − Mg = Ma
_{1}
(1)
Forces on mass m:
ΣF
_{y}
= mg − T = ma
_{2}
(2)
Accelerations balance:
a
_{2}
= 2a
_{1}
(3)
The relation between the accelerations can be found by measuring the distances that the masses travels, while mass m is moving a distance of x mass M will move a distance of x/2 so the acceleration of M is half the acceleration of m.
Solving for a
_{1}
a
_{2}
and T we get:
And the tension in the rope is:
The tension T is uniform along the cable.
The value of T
_{1}
is:
T
_{1}
= 2T
Notice that if we apply a force F downward at mass m we could lift a mass of 2F, so this system multiplies the force by 2.
Friction of two masses example - 15
▲
Find the acceleration of the system of masses neglecting the mass of the string and the inertia of the pulley.
On mass m:
ΣF
_{x}
= T cosθ − (mg − T sinθ)μ = ma cosθ
(1)
On mass M:
ΣF
_{y}
= Mg − T = Ma
(2)
The term a cosθ in equation (1) is because the acceleration of the two masses are different, the acceleration of mass m is coincide with the direction of the tension T while the actual acceleration is along the x axis.
Acceleration of M is:
Acceleration of m is:
a
_{m}
= a cosθ (in the x direction)
Friction of two masses example - 16
▲
m
_{1}
kg
m
_{2}
kg
m
_{3}
deg
a
_{1}
m/s
^{2}
a
_{2}
m/s
^{2}
a
_{3}
m/s
^{2}
T
_{1}
N
T
_{2}
N
Acceleration directions
m
_{1}
m
_{2}
m
_{3}
Find the acceleration of the system of masses neglecting the mass of the string and the mass of the pulley.
We assume arbitrary that all three accelerations are downward and are equal to a
_{1}
a
_{2}
and a
_{3}
. Now we can write the forces acting on each mass as:
ΣF
_{y}
on mass m
_{1}
m
_{1}
g − T
_{1}
= m
_{1}
a
_{1}
(1)
ΣF
_{y}
on mass m
_{2}
m
_{2}
g − T
_{2}
= m
_{2}
a
_{2}
(2)
ΣF
_{y}
on mass m
_{3}
:
m
_{3}
g − T
_{2}
= m
_{3}
a
_{3}
(3)
ΣF
_{y}
on pulley B:
T
_{1}
= 2 T
_{2}
(4)
We can see that we have 5 unknowns and only 4 equations so we have to derive another equation from the accelerations, a
_{1}
is equal to the acceleration of pulley B which sould be half the accelerations of a
_{2}
and a
_{3}
but in negative sign because when a
_{1}
is going down pulley B is going up, but we chose the accelerations in the downward direction.
Σ Accelerations:
a
_{1}
= −
(
a
_{2}
+ a
_{3}
) / 2
(5)
a
_{3}
= − 2a
_{1}
− a
_{2}
The minus sign is because a
_{1}
is in the opposite direction to a
_{2}
and a
_{3}
. Substitute the value of T
_{1}
from eq. (4) and a
_{3}
from eq. (5) to eq. (1) (2) and (3) to get the matrix form:
Solving by
Cramer's rule
we have:
D = 2 (m
_{2}
m
_{3}
+ m
_{2}
m
_{3}
) + m
_{1}
(m
_{3}
+ m
_{2}
) = m
_{1}
m
_{2}
+ m
_{1}
m
_{3}
+ 4m
_{2}
m
_{3}
After division of the numerator and the denominator by m
_{1}
m
_{2}
m
_{3}
we got another form for T
_{2}
If all masses are the same and equal to M then:
T
_{2}
= 2 g M / 3
Friction of two masses example - 17
▲
from the forces diagram we can write the equilibrium equations:
On mass m
_{1}
ΣF
_{x}
= T
_{1}
= f
_{1}
= m
_{1}
g μ
_{1}
(1)
On mass m
_{2}
ΣF
_{x}
= T
_{2}
− f
_{1}
− f
_{2}
= m
_{2}
a
T
_{2}
− m
_{1}
g μ
_{1}
− (m
_{1}
+ m
_{2}
) μ
_{2}
g = m
_{2}
a
(2)
On mass m
_{3}
ΣF
_{y}
= m
_{3}
g − T
_{2}
= m
_{3}
a
(3)
(4)
From eq.
(3)
T
_{2}
can be found
T
_{2}
=
m
_{3}
(
g − a
)
=
Q:
Given three masses m
_{1}
equal to m, mass m
_{2}
equal to 3m and m
_{3}
equal to 2m, μ
_{1}
= 0 and masses m
_{2}
and m
_{3}
are moving at a constant velocity. Find the value of μ
_{2}
, T
_{1}
and T
_{2}
S:
For steady velocity we need that a = 0
From the acceleration equation
(4)
we have:
m
_{3}
− m
_{1}
μ
_{2}
− m
_{2}
μ
_{2}
=
0
Because there is no friction between surfaces of masses m
_{1}
and m
_{2}
then: T
_{1}
= 0
Because the acceleration is 0 the tension T
_{2}
in cable m
_{3}
is: T
_{2}
= m
_{3}
g
Friction of two masses example - 18
▲
Mass M is equal to 5 kg and mass m to 3 kg the dynamic friction coefficient between the surfaces is 0.12, at time t = 0 mass M has initial velocity of 5 m/s
^{2}
find the acceleration of the system until stoppage and the distance traveled in the first two seconds.
According to the second law of motion F = m a, the total force to drive the system have to overcome the weight of mass m and the friction caused by moving mass M.
m g + μ M g = −(M + m)a
(1)
(2)
Because
m g
>
M g μ
then the system will finally stop moving to the left and mass m will start to fall so v
_{t}
= 0
From acceleration equations we have:
v
_{t}
^{2}
= v
_{0}
^{2}
+ 2aS
And the distance until stoppage (v
_{t}
= 0):
The time traveled as a function of acceleration and v
_{0}
and remembering that v
_{t}
= 0 can be calculated from:
v
_{t}
= v
_{0}
+ a t
(3)
Substitute the value of a we get the total travel time until stoppage:
The acceleration until stoppage according to eq. (2) is:
S
_{(2s)}
= v
_{0}
t + a t
^{2}
/ 2
= 5 * 2 − 4.4 * 4 / 2 = 1.2 m
Friction of two masses example - 19
▲
A force F is applied to mass m
_{2}
which is sliding on a frictionless wheels. Mass m
_{1}
is located on mass m
_{2}
with a friction coefficient of μ
_{1}
this nass is connected to mass m
_{3}
by a frictionless pulley, the friction coefficient between surfaces m
_{2}
and m
_{3}
is μ
_{2}
. Find the minimum force F that will keep mass m
_{1}
in place and the acceleration of the masses.
Friction force f
_{2}
is effective only when the system is accelerates to the right:
f
_{2}
= m
_{3}
a μ
_{2}
(1)
In order to keep the system of masses m
_{1}
and m
_{3}
at rest we need that:
m
_{1}
a = m
_{3}
g
(2)
(in this case f
_{1}
= f
_{2}
= 0 because there is no movement relative to mass m
_{2}
)
According to second law of motion:
F = (m
_{1}
+ m
_{2}
+ m
_{3}
) a
(3)
From eq. (2)
Substitute a into eq. (3)
Question:
Given three masses m
_{1}
= 10 kg, m
_{2}
= 2 kg and m
_{3}
= 1 kg. Find the minimum force F that will keep masses m
_{1}
and m
_{2}
in place and the acceleration of the masses if the force F is twice as the force found.
Solution:
For steady velocity we need that a = 0
From the acceleration equation we have: m
_{3}
− m
_{2}
μ
_{2}
+ m
_{1}
μ
_{2}
= 0
Because there is no friction between surfaces of masses m
_{1}
and m
_{2}
then: T
_{1}
= 0
Because the acceleration is 0 the tension T
_{2}
in cable m
_{3}
is: T
_{2}
= m
_{3}
g
Friction of two masses example - 20
▲
Find the acceleration of the system of masses neglecting the mass of the string and the inertia of the pulley.
The friction force value is:
f = μ m
_{1}
g cos θ
When mass m
_{2}
is moving a distance of x meters the mass m
_{1}
will travel half that distance hence the acceleration will be half the acceleration of m
_{2}
eq. (3).
From equilibrium of forces we have:
On mass m
_{1}
:
ΣF
_{x}
= T
_{1}
− m
_{1}
g sinθ − μm
_{1}
g cos θ = m
_{1}
a
_{1}
(1)
On mass m
_{2}
:
ΣF
_{y}
= m
_{2}
g − T
_{2}
= m
_{2}
a
_{2}
(2)
Accelerations:
a
_{2}
= 2a
_{1}
(3)
From tensions:
T
_{2}
= 2T
_{1}
(4)
Friction of two masses example - 21
▲
Three masses are hanging on frictionless pulleys as shown in the figure. Find the acceleration of the masses.
We choose arbitrary the direction of the accelerations of masses m
_{1}
and mass m
_{2}
as upward and the acceleration of m
_{2}
downward.
From the forces diagram on the masses we have.
On mass m
_{1}
ΣF
_{y}
= T − m
_{1}
g = m
_{1}
a
_{1}
(1)
On mass m
_{2}
:
ΣF
_{y}
= m
_{2}
g − 2T = m
_{2}
a
_{2}
(2)
On mass m
_{3}
:
ΣF
_{y}
= T − m
_{3}
g = m
_{3}
a
_{3}
(3)
From accelerations:
(4)
Note: when mass m
_{1}
is moving x
_{1}
distance upward and mass m
_{3}
a distance of x
_{3}
upward, the contribution to the displacement of mass m
_{2}
is the sum of half of this displacements x
_{2}
= (x
_{1}
+ x
_{3}
) /
2
in the downward direction.
The equations with the unknowns T, a
_{1}
, a
_{2}
and a
_{3}
can be solved by
Cramer's rule
or by direct substitution.
Write the equations in the matrix form
D = − m
_{2}
m
_{3}
− 4m
_{1}
m
_{3}
− m
_{1}
m
_{2}
= − (m
_{1}
m
_{2}
+ m
_{2}
m
_{3}
+ 4m
_{1}
m
_{3}
)
Find the acceletations and the tension in the rope if all the masses are equal to m
_{1}
= m
_{2}
= m
_{3}
= m
T = 2 m g / 3
a
_{1}
= − g / 3
a
_{2}
= − g / 3
a
_{3}
= − g / 3
T = 2 m / 3
Notice
that if m
_{1}
= m
_{3}
= m and m
_{2}
= 2m then the accelerations will be 0 and T = m
If the result of the acceleration is negative then the motion is opposite to that chosen.
Friction of two masses example - 22
▲
Mass m
_{1}
is located on a slope with an angle θ and friction coefficient of μ, the mass ia connected to a mass m
_{2}
and to the wall with two pullies as shown. Find thq biggest value of m
_{2}
that the system will be at rest, and the accelerations of the bodies.
The friction force value is:
f = μ m
_{1}
g cos θ
When mass m
_{2}
is moving a distance of x meters the mass m
_{1}
will travel half that distance hence the acceleration will be half the acceleration of m
_{2}
eq. (3).
From equilibrium of forces we have:
On mass m
_{1}
:
ΣF
_{x}
= T
_{1}
− m
_{1}
g sinθ − μm
_{1}
g cos θ = m
_{1}
a
_{1}
(1)
On mass m
_{2}
:
ΣF
_{y}
= m
_{2}
g − T
_{2}
= m
_{2}
a
_{2}
(2)
Accelerations:
a
_{2}
= 2a
_{1}
(3)
From tensions:
T
_{1}
= 2T
_{2}
(4)
Substituting eq. (3) and (4) into eq. (1) and (2) and solving for a
_{1}
and T
_{2}
we get:
Friction of two masses example - 23
▲
Two pulleys A and B are connected as shown in the figure, the cable which holds pulleys A and B is connected to mass m
_{1}
with an angle θ and is resting on a surface with friction coefficient of μ, pulley B is connected to mass m
_{2}
. Find the minimum weight of mass m
_{2}
that will cause mass m
_{1}
to move and the tension of the cable in this case, also find the equivalent and the direction of the force acting on pulley A.
The friction force acting on mass m
_{1}
is:
f = (m
_{1}
g − T
_{1}
sin θ) μ
From the forces diagram on mass m
_{1}
and m
_{2}
and assuming that the acceleration is to the downward direction on mass m
_{2}
we have:
On mass m
_{1}
ΣF
_{x}
= T
_{1}
cosθ − (m
_{1}
g − T
_{1}
sin θ) μ = m
_{1}
a
_{1}
(1)
On mass m
_{2}
:
ΣF
_{y}
= m
_{2}
g − T
_{2}
= m
_{2}
a
_{2}
(2)
From pulley B:
T
_{2}
= 2T
_{1}
(3)
Because at rest the system acceleration equals 0 then a
_{1}
= a
_{2}
= 0
From eq. (1)
T
_{1}
cosθ − (m
_{1}
g − T
_{1}
sin θ) μ = 0
(1a)
From eq. (2) and (3)
T
_{2}
= m
_{2}
g → T
_{1}
= m
_{2}
g / 2
(1b)
Substitute eq. (1b) into (1a) to get the value of mass m
_{2}
Substitute eq. (1b) into (1a) to get the value of mass m
_{2}
And the tension T
_{1}
in the cable is:
In order to find the value and the direction of the force acting on pulley A according to the force diagram
Friction of two masses example - 24
▲
Three masses are connected with a rope of 0 mass as shown in the figure, a force F is apply to mass m
_{3}
, friction coefficient μ if present between mass m
_{1}
and the surface that is tilted by θ deg. Find the value of the force F that will keep the system at rest, if the force F is cancelled find the acceleration of the system.
In order to determine the direction of the motion we have to analyze the relationship of the equations:
m
_{3}
>
m
_{2}
+ m
_{1}
sinθ
Motion is m
_{3}
down
(a)
m
_{3}
<
m
_{2}
+ m
_{1}
sinθ
Motion is m
_{3}
up
(b)
m
_{3}
=
m
_{2}
+ m
_{1}
sinθ
No motion
(c)
Assume that the acceleration of mass m
_{3}
is downward then:
On mass m
_{1}
ΣF
_{x}
= T
_{1}
− μ m
_{1}
g cos θ − m
_{1}
g sinθ = m
_{1}
a
(1)
On mass m
_{2}
ΣF
_{y}
= T
_{2}
− m
_{2}
g − T
_{1}
= m
_{2}
a
(2)
On mass m
_{3}
ΣF
_{y}
= m
_{3}
g + F − T
_{2}
= m
_{3}
a
(3)
The same equations when the acceleration is to the left are:
On mass m
_{1}
ΣF
_{x}
= m
_{1}
g sinθ − T
_{1}
− μ m
_{1}
g cosθ = m
_{1}
a
(4)
On mass m
_{2}
ΣF
_{y}
= m
_{2}
g + T
_{1}
− T
_{2}
= m
_{2}
a
(5)
On mass m
_{3}
ΣF
_{y}
= T
_{2}
− m
_{3}
g − F = m
_{3}
a
(6)
We got three equations with three unknowns T
_{1}
T
_{2}
and a.
Solving eq.
(1) (2)
and
(3)
according to
Cramer's rule
we get:
Solving eq.
(4) (5)
and
(6)
according to
Cramer's rule
we get:
The minimum force F nedded to apply when a = 0, when the motion is intended to be to the left is:
F = g ( m
_{1}
sinθ + m
_{2}
− m
_{3}
) − m
_{1}
μ
_{s}
g cosθ
The maximum force F nedded to apply when a = 0, when the motion is intended to be to the right is:
F = g ( m
_{1}
sinθ + m
_{2}
− m
_{3}
) + m
_{1}
μ
_{s}
g cosθ
In the case when F = 0 mass m
_{3}
that keeps the system at rest is in the range of:
m
_{1}
(
sinθ − μ cosθ
) +
m
_{2}
<
m
_{3}
<
m
_{1}
(
sinθ + μ cosθ
) +
m
_{2}
If all three masses are of the same weight M then the force F at rest should be in the range:
M g
(
sinθ − μ
_{s}
cosθ
)
<
F
<
M g
(
sinθ
+
μ
_{s}
cosθ
)
Friction of two masses example - 25
▲
Two masses m
_{1}
and m
_{2}
are connected throught 2 pulleys as shown in the figure the friction of m
_{1}
and the surface is μ. Find the maximum and minimum values of m
_{2}
that will steel keep the system at rest and the acceleration of m
_{2}
when m
_{2}
= 3m
_{1}
the friction coefficient is μ = 0.1.
we assume that the acceleration of mass m
_{1}
is upward along the slope, then the friction force direction will be downward as shown in the force diagram at right.
The friction force is equal to:
f = m
_{1}
g cosθ μ
From the forces diagram on masses m and M we get the forces equations:
On mass m
_{1}
ΣF
_{x}
= T
_{1}
− m
_{1}
g sinθ − m
_{1}
g cosθ μ = m
_{1}
a
_{1}
(1)
On mass m
_{2}
:
ΣF
_{y}
= m
_{2}
g − T
_{2}
= m
_{2}
a
_{2}
(2)
From the pulley:
T
_{2}
= 2T
_{1}
(3)
From
accelerations:
a
_{1}
= 2a
_{2}
(4)
Solving for T
_{1}
we get:
And the acceleration is:
(need that a
_{2}
> 0)
Mass m
_{2}
is:
At rest a = 0 and mass m
_{2}
is:
m
_{2}
= 2 m
_{1}
( sinθ + μ cosθ )
(maximum value)
Note:
If the masses are moving at a constant speed then the acceleration is 0 and according to the equation
v
_{t}
= v
_{0}
+ a t
we can see that the bodies will keep moving at the speed of the initial velocity
v
_{t}
= v
_{0}
.
If m
_{2}
is bellow the minimum value then mass m
_{1}
will slip down the slope and the friction force f will be upward as shown at left.
On mass m
_{1}
ΣF
_{x}
= m
_{1}
g sinθ − T
_{1}
− m
_{1}
g cosθ μ = m
_{1}
a
_{1}
(5)
On mass m
_{2}
:
ΣF
_{y}
= T
_{2}
− m
_{2}
g = m
_{2}
a
_{2}
(6)
From the pulley:
T
_{2}
= 2T
_{1}
(3)
From
accelerations:
a
_{1}
= 2a
_{2}
(4)
Solving for T
_{1}
we get:
And the acceleration is:
(need that a
_{2}
> 0)
And the mass m
_{2}
is:
At rest a = 0 and mass m
_{2}
is:
m
_{2}
= 2 m
_{1}
( sinθ − μ cosθ )
(minimum value)
Because there is friction in the system in order to find the acceleration of mass m
_{2}
when m
_{2}
= 3m
_{1}
we have to determine the direction of the movment according to the following conditions.
(mass m
_{2}
up)
2m
_{1}
(sinθ − μ cosθ) > m
_{2}
> 2m
_{1}
(sinθ + μ cosθ)
(mass m
_{2}
down)
Friction of two masses example - 26
▲
Masses m
_{1}
and m
_{2}
are connected with two pullys as shown in the left figure, the friction coefficient of mass m
_{1}
and the surface is μ find the acceleration of the masses and the tension in the cable connected to mass m
_{1}
.
The acceleration direction of mass m
_{2}
will be downward.
From the forces diagram we have.
On mass m
_{1}
ΣF
_{x}
= T
_{1}
− m
_{1}
g μ = m
_{1}
a
_{1}
(1)
On pulley A:
ΣF
_{y}
= 2 T
_{1}
= T
_{2}
(2)
a
_{1}
= 2 a
_{2}
(3)
On pulley B:
ΣF
_{y}
= m
_{2}
g − 2 T
_{2}
= m
_{2}
a
_{3}
(4)
a
_{2}
= 2 a
_{3}
(5)
We get 5 simple equations with 5 unknowns T
_{1}
T
_{2}
a
_{1}
a
_{2}
and a
_{3}
.
Solving for a
_{3}
we get:
a
_{1}
can be found by eq. (3) and (5)
a
_{1}
= 2a
_{2}
= 4a
_{3}
Condition for movment (a
_{3}
> 0):
m
_{2}
> 4m
_{1}
μ
And the tension T
_{1}
from eq. (1) is:
Friction of two masses example - 27
▲
Mass m
_{1}
is located on a slope with an angle of θ degree and a friction coefficient of μ, mass m
_{2}
ia accelerating to the left direction on a frictionless wheels. Find the value of the acceleration that will keep mass m
_{1}
at rest.
Because the force acting on mass m
_{2}
is to the left according to Newtons law of motion mass m
_{1}
produce a force opposite to the force on m
_{2}
.
The friction force value is:
f = N μ = (m
_{1}
g cosθ + m
_{1}
a sinθ) μ
From equilibrium of forces in the slope direction we have:
ΣF
_{x}
= m
_{1}
g sinθ + (m
_{1}
g cosθ + m
_{1}
a sinθ) μ = m
_{1}
a cosθ
And the acceleration is:
Friction of two masses example - 28
▲
A force F cause masses m
_{1}
and m
_{2}
to accelerate. mass m
_{2}
is attached to mass m
_{1}
only by friction force. Find the acceleration needed to prevent mass m
_{2}
to slip down.
From the forces diagram on mass m
_{2}
the friction force should be equal or more then the weight of mass m
_{2}
in order not to slip, the friction force is a function of the acceleration and is equal to:
f = m
_{2}
a μ
_{s}
ΣF
_{y}
= m
_{2}
a μ
_{s}
≥ m
_{2}
g
The forces acting on mass m
_{2}
are:
ΣF
_{x}
= F − (m
_{1}
+ m
_{2}
)gμ
_{2}
= (m
_{1}
+ m
_{2}
)a
Eliminating the value of F and substitute a = gμ
_{1}
we get:
Friction of two masses example - 29
▲
Mass m is hanging on pulley A and is connected to two other pullys as shown in the figure, find a) the force F needed to keep mass m at rest b) the acceleration of the force downward and c) suppose that mass m ia the mass of a person lifting himselef what should be the force F that will keep him in place.
We choose the acceleration direction to cope with the direction of force F.
a) From the free body diagram of the pulleys we have.
On pulley A
ΣF
_{y}
= 2 T
_{1}
= m g
(1)
On pulley B:
ΣF
_{y}
= 2 F = T
_{1}
(2)
On pulley C:
ΣF
_{y}
= 2 F = T
_{2}
(3)
From eq. (1) and (2) we get that: F = m g / 4 or m g = 4 F and T
_{1}
= T
_{2}
b) In order to find the acceleration of force F relative to the acceleration of mass m we have to analyze the distances that each pulley travels.
Friction of two masses example - 30
▲
A weight of 70 kgf is connected with three cables to a system of pulleys as shown in the left figure. Find the tension in each cable holding the weight.
We choose the tension of the right cable to be T and the forces diagram is drawn in the figure at right
The weigh is supported by three cables with a total tension of
2T + T / 2 + T = 3.5 T
.
From equilibrium of the weight we have:
W = 3.5 T
T = W / 3.5 = 70 / 3.5 = 20N
And the tension in the cables are:
2T = 2* 20 = 40N
T / 2 = 20 / 2 = 10N
T = 20N
Friction of two masses example - 31
▲
Verify the minimum force F needed to lift a weight of 100 kgf if the weight is connected by the series of pulleys as shown in the left figure.
At the right figure the tension in each cable started at the weight W is marked, it can easilly seen that each pulley is reducing the force F by half (n pulleys gives F = W / 2
^{n}
).
From equilibrium we have:
F = W / 16 = 100 / 16 = 6.25 kgf
Friction of two masses example - 32
▲
Three bodies are connected through a pulley as shown in the left figure, if the friction between the surface and mass m
_{1}
is μ
_{s}
find the accelerations of the bodies.
We will mark the acceleration of mass m
_{2}
and m
_{3}
as a
_{2}
this acceleration is the acceleration relative to the pulley. Therfore the real accelerations will be:
a
_{m2}
= a
_{2}
− a
_{1}
and
a
_{m3}
= a
_{1}
+ a
_{2}
The kinetic friction force acting on mass m
_{1}
is:
f = m
_{1}
g μ
From the free body diagram on mass m
_{1}
and assuming that the acceleration is to the right a
_{1}
:
On mass m
_{1}
ΣF
_{x}
= T
_{1}
− f = m
_{1}
a
_{1}
(1)
On mass m
_{2}
:
ΣF
_{y}
= T
_{2}
− m
_{2}
g = m
_{2}
a
_{m2}
(2)
On mass m
_{3}
:
ΣF
_{y}
= m
_{3}
g − T
_{2}
= m
_{3}
a
_{m3}
(3)
On the pulley:
T
_{1}
= 2T
_{2}
(4)
After substituting eq. (4) and the values for a
_{m2}
and a
_{m3}
we get the system of equations:
Solving this system of linear equations the determinant is equal to: D = 4 m
_{2}
m
_{3}
+ m
_{1}
( m
_{3}
+ m
_{2}
)
And by
Cramer's rule
we get:
Friction of two masses example - 33
▲
Three masses are connected to each other by ropes of 0 mass, a force F is apply to mass m
_{3}
. Find the value of the tension in the ropes and the acceleration of the masses if a friction coefficient of μ exists between the masses and the surface.
From the free body diagram the friction forces and the equilibrium equations of the masses are:
Friction forces:
f
_{1}
= m
_{1}
g μ
f
_{2}
= m
_{2}
g μ
f
_{3}
= m
_{3}
g μ
ΣF
_{x}
= T
_{1}
− m
_{1}
g μ = m
_{1}
a
(1)
ΣF
_{x}
= T
_{2}
− T
_{1}
− m
_{2}
g μ = m
_{2}
a
(2)
ΣF
_{x}
= F − T
_{2}
− m
_{3}
g μ = m
_{3}
a
(3)
We got three equations with three unknowns T
_{1}
T
_{2}
and a
According to
Cramer's rule
the value of the coefficients determinant is
D = − (m
_{1}
+ m
_{2}
+ m
_{3}
)
Another way to solve the problem is to look on the three masses as one mass equal to M = m
_{1}
+ m
_{2}
+ m
_{3}
From the free body diagram we have:
F − M g μ = M a
and a is:
Once we found the acceleration a we can use eq. (1) and (2) to find T
_{1}
and T
_{2}
Friction of two masses example - 34
▲
Two masses are connected by a rope of 0 mass to each other, the masses are placed on a slope of angle θ the friction coefficient between the masses and the surface are μ
_{1}
and μ
_{2}
respectively. Find a) the acceleration and the tension in the rope, b) the maximum angle that the masses will stay at rest.
Free body diagram:
Friction forces caused by the masses movment are:
f
_{1}
= m
_{1}
g μ
_{1}
f
_{2}
= m
_{2}
g μ
_{2}
From the free body diagram write the equilibrium equations of each mass:
ΣF
_{x}
= m
_{1}
g sinθ − T − m
_{1}
g cosθ μ
_{1}
= m
_{1}
a
(1)
ΣF
_{x}
= T + m
_{2}
g sin θ − m
_{2}
g cosθ μ
_{2}
= m
_{2}
a
(2)
We got two equations with two unknowns T and a. Solving by
Cramer's rule
we have:
D = −(m
_{1}
+ m
_{2}
)
The maximum angle for rest condition can be calculated by assuming that a = 0 and we get:
(3)
Note
: In order to create tension in the rope we need that the lower mass acceleration will be bigger then the upper mass acceleration so we get:
m
_{1}
g sinθ − m
_{1}
g μ
_{1}
cosθ = m
_{1}
a
_{m1}
➞
a
_{m1}
= g(sinθ − μ
_{1}
cosθ)
m
_{2}
g sinθ − m
_{2}
g μ
_{2}
cosθ = m
_{2}
a
_{m2}
➞
a
_{m2}
= g(sinθ − μ
_{2}
cosθ)
From both equations it is clear that the mass with lower friction coefficient will have bigger acceleration and should be in the lower position.
Given two masses of 5 kg and 10 kg and a friction coefficients of 0.1 and 0.2 respectively. Find a) the order of the location of the masses that will cause a tension in the rope b) the minimum angle that will cause the masses to slipe and c) the acceleration of the masses if the angle is 30 degree.
a) From the privious note the lower mass should be the mass with lower friction coefficient m
_{1}
= 5 kg
b) from eq (3) we can find the lowest angle that the masses will slipe (ΣF
_{x}
= 0).
c) if the slope angle is 30 degree then the ecceleration and the tension as found before will be:
Friction of two masses example - 35
▲
Mass m
_{1}
is located on mass m
_{2}
and has a friction coefficients of μ
_{1S}
and μ
_{1k}
the friction coefficients of m
_{2}
and the lower surface are μ
_{2S}
and μ
_{2k}
. Find a) the biggest force F that can be applied to m
_{1}
without causing mass m
_{1}
to slip, b) if the applied force is twice the force found what are the accelerations of the masses.
There are 4 movements posibillitys for the masses:
1) No movement at all, this happens when:
F < m
_{1}
g μ
_{1S}
and
m
_{1}
μ
_{1S}
< (m
_{1}
+ m
_{2}
) μ
_{2S}
(f
_{1}
< f
_{2}
)
2) Only m
_{1}
is moving, this happens when:
F > m
_{1}
g μ
_{1S}
and
m
_{1}
μ
_{1S}
< (m
_{1}
+ m
_{2}
) μ
_{2S}
(f
_{1}
< f
_{2}
)
The acceleration of m
_{1}
is:
3) m
_{1}
and m
_{2}
are moving together, this happens when:
F > m
_{1}
g μ
_{1S}
and
m
_{1}
μ
_{1S}
= (m
_{1}
+ m
_{2}
) μ
_{2S}
(f
_{1}
= f
_{2}
)
The acceleration of m
_{1}
is:
The maximum force F that the masses will move together is
4) m
_{1}
and m
_{2}
moves at different accelerations, this happens when:
F > m
_{1}
g μ
_{1S}
and
m
_{1}
μ
_{1S}
> (m
_{1}
+ m
_{2}
) μ
_{2S}
(f
_{1}
> f
_{2}
)
The acceleration of m
_{1}
is:
The maximum friction forces acting on m
_{1}
is
f
_{1}
= m
_{1}
g μ
_{1S}
From the free body diagram mass m
_{1}
will move when force F overcomes the static friction forc f
_{1}
.
ΣF
_{x}
= F − m
_{1}
g μ
_{1S}
= m
_{1}
a
_{1}
(1)
Once the force F is bigger then f
_{1}
m
_{1}
accelerates by:
ΣF
_{x}
= m
_{1}
g μ
_{1}
− (m
_{1}
+ m
_{2}
) g μ
_{2}
= m
_{2}
a
_{2}
(2)
We got two equations with two unknowns a
_{1}
and a
_{2}
Note:
both accelerations must be positive should be in the direction of the force F so a
_{1}
>= 0 and a
_{2}
>= 0
a) In order to keep both masses moving together we have to fullfil the condition that a
_{1}
= a
_{2}
b) if the force is twice the force found in a) and is 2F then the accelerations of the bottom mass will be the same because the force acting on mass m
_{2}
is only the friction force. The acceleration on mass m
_{1}
will be:
Q.
A masse of 3 kg is mounted on top of a mass of 7 kg the friction coefficients between the masses are
μ
_{1S}
= 0.8 and μ
_{1k}
= 0.1 and between the lower mass and the surface are μ
_{2S}
= 0.2 and μ
_{2k}
= 0.1. Find the maximum force F that will pul both masses together.
S.
Friction of two masses example - 35
▲
Mass m
_{1}
is located on mass m
_{2}
and has a friction coefficients of μ
_{1S}
and μ
_{1k}
the friction coefficients of m
_{2}
and the lower surface are μ
_{2S}
and μ
_{2k}
. Find a) the biggest force F that can be applied to m
_{1}
without causing mass m
_{1}
to slip, b) if the applied force is twice the force found what are the accelerations of the masses.
The maximum friction forces acting on m
_{1}
is
f
_{1}
= m
_{1}
g μ
_{1S}
From the free body diagram mass m
_{1}
will move when force F overcomes the static friction forc f
_{1}
.
ΣF
_{x}
= F − m
_{1}
g μ
_{1S}
= m
_{1}
a
_{1}
(1)
Once the force F is bigger then f
_{1}
m
_{1}
accelerates by:
ΣF
_{x}
= m
_{1}
g μ
_{1}
− (m
_{1}
+ m
_{2}
) g μ
_{2}
= m
_{2}
a
_{2}
(2)
We got two equations with two unknowns a
_{1}
and a
_{2}
Note:
both accelerations must be positive should be in the direction of the force F so a
_{1}
>= 0 and a
_{2}
>= 0
a) In order to keep both masses moving together we have to fullfil the condition that a
_{1}
= a
_{2}
b) if the force is twice the force found in a) and is 2F then the accelerations of the bottom mass will be the same because the force acting on mass m
_{2}
is only the friction force. The acceleration on mass m
_{1}
will be:
Friction of two masses example - 36
▲
Mass m
_{1}
is moving at an initial speed of v
_{0}
on mass m
_{2}
the kinetic friction coefficient between mass m
_{1}
and mass m
_{2}
is μ
_{1}
and between mass m
_{2}
and the surface is μ
_{2}
. What is the distance travelled by mass m
_{1}
relative to mass m
_{2}
until mass m
_{1}
stopes.
The friction force f
_{1}
= m
_{1}
g μ
_{1}
is causing mass m
_{1}
to decelerate from the initial velocity v
_{0}
but this force is causing mass m
_{2}
to accelerate after overcomming the friction force f
_{2}
= (m
_{1}
+ m
_{2}
) g μ
_{2}
. The relative acceleration of mass m
_{1}
relative to mass m
_{2}
is
a
_{1}
− a
_{2}
From the free body diagram the equilibrium equations are:
ΣF
_{x}
= m
_{1}
g μ
_{1}
= m
_{1}
(a
_{1}
− a
_{2}
)
(1)
ΣF
_{x}
= m
_{1}
g μ
_{1}
− (m
_{1}
+ m
_{2}
) g μ
_{2}
= m
_{2}
a
_{2}
(2)
We got two equations with two unknowns a
_{1}
and a
_{2}
From eq. (2)
From eq. (1)
The final velocity is v
_{t}
= 0 and we find the total travel distance by:
v
_{t}
^{2}
= v
_{0}
^{2}
− 2 a D
The minus sign is because deceleration is negative acceleration.
Friction of two masses example - 37
▲
Two masses m
_{1}
and m
_{2}
are located on the floor the masses then conected by a pulley as shown in the figure. Find the acceleration of the masses and the tension in the rope connecting the masses if a force of F is apply to the pulley upward, suppose that m
_{1}
> m
_{2}
.
We mark the accelerations of the masses as a
_{1}
and a
_{2}
, from the free body diagram we get:
On mass m
_{1}
:
ΣF
_{y}
= T − m
_{1}
g = m
_{1}
a
_{1}
(1)
On mass m
_{2}
:
ΣF
_{y}
= T − m
_{2}
g = m
_{2}
a
_{2}
(2)
On the pulley:
ΣF
_{y}
= F = 2 T
(3)
From eq.
(3)
we get:
T = F / 2
And substituting T to eq.
(1)
and
(2)
we get:
Notes:
because accelerations a
_{1}
and a
_{2}
can not be negative (opposite to the direction shown) because of the floor there are some unique casses:
Samll mass m
_{2}
F < 2 m
_{2}
g
The force F is not enough to lift any one of the masses
Bigger mass m
_{1}
2 m
_{2}
g < F < 2 m
_{1}
g
The force F will lift the small mass but not the bigger mass
2 m
_{1}
g < F
The force F will lift both masses
For example if mass m
_{1}
= 30 kg and m
_{2}
= 20 kg then find the accelerations when F = 100N, F = 400N and F = 1000N.
F
[N]
a
_{1}
[m/s
^{2}
]
a
_{2}
[m/s
^{2}
]
Notes
100N
100 / 30 − 9.8 = − 6.5
100 / 20 − 9.8 = − 4.8
No lift of any mass a
_{1}
= 0 a
_{2}
= 0
200N
200 / 30 − 9.8 = − 3.1
200 / 20 − 9.8 = 0.2
Only small mass is lifted a
_{1}
= 0
1000N
400 / 30 − 9.8 = 3.5
400 / 20 − 9.8 = 10.2
Both masses moves upward
Friction of two masses a force and pulley - example 38
▲
m
_{1}
kg
m
_{2}
kg
μ
F
N
θ
deg
a
m/s
_{2}
T
N
f
N
Mass m
_{1}
and m
_{2}
are at rest when a force F is applied at an angle of θ degree. Find the maximum force F that the system will stay at rest and the acceleration of the system.
The friction force acting on m
_{1}
is:
f = N μ = (m
_{1}
g + F sin θ) μ
.
In order to determine the possible direction of the motion we have to check the forces acting on mass m
_{1}
without friction force.
if
F cosθ
>
m
_{2}
g
m
_{1}
moves to the left
(a)
if
F cosθ
<
m
_{2}
g
m
_{1}
moves to the right
(b)
After we determined the direction of the motion we can set the right direction of the friction force and check if there is motion at all due to the friction force.
Let assume that the acceleration of m
_{1}
is to the left, from the free body diagram we have
On m
_{1}
F cosθ − f − T = m
_{1}
a
(1)
On m
_{2}
T − m
_{2}
g = m
_{2}
a
(2)
From eq. (1) and (2) we can find the values of a and T
If we asumme that the acceleration of mass m
_{1}
is to the right direction the equilibrium equations will be as follows:
On m
_{1}
T − F cosθ − f = m
_{1}
a
(3)
On m
_{2}
m
_{2}
g − T = m
_{2}
a
(4)
If we get a negative value for the acceleration in both directions then motion is not possible.
The conditions for motion are:
If
F cosθ − f
>
m
_{2}
g
m
_{1}
moves to the left
(c)
If
F cosθ + f
<
m
_{2}
g
m
_{1}
moves to the right
(d)
The range of the force F that the system is at rest is when it is equal to the friction force in both directions:
the static and dynamic friction coefficient between the surface and m
_{1}
is μ = 0.4. Find the friction force acting on m
_{1}
if the force equal 250N and θ = 30 degree and m
_{1}
= 10kg and m
_{2}
= 30 kg.
First we check conditions (c) and (d) to verify if the masses are moving.
Condition
(c)
30 * g > 250 * cos30 + (10 * g + 250 * sin30) 0.4
False
Condition
(d)
30 * g < 250 * cos30 - (10 * g + 250 * sin30) 0.4
False
Because both conditions are false the system is at rest.
a) Friction force can not be calculated by the equation for f because there is no movement, f will be calculated by the difference between force of mass m
_{2}
g and F * cosθ
f = m
_{2}
g − F * cosθ
= 30 * 9.8 − 250 * cos30 = 77.5 N
b) maximum force F
F = g(10*0.4 + 30) /(cos 30 - sin 30 * 0.4) = 500.3 N
c) acceleration when F = 0
a = −g(10*0.4 - 30) /(10 + 30) = 6.4 m / s
^{2}
Friction example - 39
▲
Two masses m
_{1}
and m
_{2}
are located on a tilted slope of angle θ and with friction coefficient of μ. A force F is applied to the lower mass in the upward direction. Find the accelerations of the masses and the force that exerts by the masses on each other.
We will choose a positive acceleration upward.
The friction forces are:
f
_{1}
= m
_{1}
g μ cosθ
f
_{2}
= m
_{2}
g μ cosθ
From the forces diagram we can derive the equilibrium equations:
ΣF
_{x}
= F − P − f
_{1}
− m
_{1}
g sin θ = m
_{1}
a
(1)
ΣF
_{y}
= P − f
_{2}
− m
_{2}
g sin θ = m
_{2}
a
(2)
From eq. (1) and (2) we ca evaluate the values of a and P :
What will be the result of the acceleration and the force P if we change the locations of m
_{1}
with m
_{2}
and neglecting the friction assume that m
_{1}
= m and m
_{2}
= 3m and the force F = 4mg and the slope angle is 30 degree.
From the free body diagram the equilibrium equations of the bodies are:
ΣF
_{y}
= P − f
_{1}
− m
_{1}
g sin θ = m
_{1}
a
(2)
ΣF
_{x}
= F − P − f
_{2}
− m
_{2}
g sin θ = m
_{2}
a
(1)
Two masses and two pulleys example - 40
▲
Two masses m
_{1}
and m
_{2}
are connected through 2 pulleys as shown in the figure the friction of m
_{1}
and the surface is μ. Find the accelerations and the tensions in the ropes.
In order to find the direction of the motion we will analyze the forces on the pulley by neglecting the friction force, it is easy to see that if:
m
_{1}
sinθ
>
2 m
_{2}
m
_{1}
is moving downward
(a)
m
_{1}
sinθ
<
2 m
_{2}
m
_{1}
is moving upward
(b)
m
_{1}
sinθ
=
2 m
_{2}
There will be no motion
(c)
If m
_{2}
exceeds a maximum value then mass m
_{2}
will start to move down and the friction force direction will be as shown in the free body diagram at right.
The friction force is equal to:
f = N
_{1}
μ = m
_{1}
g cosθ μ
From the forces diagram on mass m
_{1}
and m
_{2}
and assuming that the acceleration is to the downward direction of mass m
_{2}
we have.
On mass m
_{1}
ΣF
_{x}
= T
_{1}
− f − m
_{1}
g sinθ = m
_{1}
a
_{1}
(1)
On mass m
_{2}
:
ΣF
_{y}
= m
_{2}
g − T
_{2}
= m
_{2}
a
_{2}
(2)
From the accelerations:
a
_{2}
= 2a
_{1}
(3)
From the pulley tension:
T
_{1}
= 2T
_{2}
(4)
Solving for a
_{1}
and T
_{2}
we get:
If we solve the case that m
_{1}
accelerates downward then the forces equations will be:
On mass m
_{1}
ΣF
_{x}
= m
_{1}
g sinθ − T
_{1}
− f = m
_{1}
a
_{1}
(5)
On mass m
_{2}
:
ΣF
_{y}
= T
_{2}
− m
_{2}
g = m
_{2}
a
_{2}
(6)
Eq. (3) and (4) are the same in both cases, solving the equations we get:
The range of m
_{2}
that the system will stay at rest is (in this case the accelerations a
_{1}
are equal to 0):
Friction of two masses example - 41
▲
Three masses are located as shown in the figure, the static friction between mass m
_{1}
and m
_{2}
is μ
_{s}
and the kinetic friction of all surfaces are μ
_{k}
. Find a) the maximum acceleration of mass m
_{2}
that will keep m
_{1}
not to slip, b) what is the biggest weight of m
_{3}
that will keep m
_{1}
from slipping and c) if the mass of m
_{3}
is twice the mass found at b) what is the accelerations of the masses.
a) the maximum force caused by the acceleration of mass m
_{1}
should be balanced by the static friction
force:
m
_{1}
a = f
_{1}
= m
_{1}
g μ
_{s}
and the maximum acceleration is:
a = g μ
_{s}
b) from the forces diagram we can write the equilibrium equations of the masses:
On mass m
_{2}
ΣF
_{x}
= T − f
_{1}
− f
_{2}
= (m
_{1}
+ m
_{2}
) a
(1)
On mass m
_{3}
ΣF
_{x}
= m
_{3}
g − T = m
_{3}
a
(2)
Substitute the acceleration found in a) and solving for m
_{3}
we get:
c) twice the mass is 2m
_{3}
and solving for a we get:
a(max) = 0.6g = 5.88 m/s
^{2}
m
_{3}
= 4.66m a = 0.36g = 3.54 m/s
^{2}
Friction of two masses example - 42
▲
Two masses are conected with an inextensible cord as shown in the figure. If mass M is 50 kg and mass m is 10 kg find the distance that traveled by the masses in 3 sec if released from rest.
We will take the x axis for both masses as parallel to the surface to make the equations more simple.
The normal forces are:
N
_{1}
= M g cosα
N
_{2}
= m g cosθ
Friction forces are:
f
_{M}
= N
_{1}
μ = M g cosα μ
f
_{m}
= N
_{2}
μ = m g cosθ μ
From the forces diagram on mass M and m and setting acceleration downward on mass m we get:
On mass M
T + M g sinα − f
_{M}
= M a
(1)
On mass m
m g cosθ − T − f
_{m}
= m a
(2)
From eq. (1) and (2):
Friction of two masses example - 43
▲
A cart of mass M is released from rest and is moving downward without friction on a surface tilted by an angle of θ degree. Mass m is located horizontally on the cart surface and has a friction coefficient of μ
_{s}
and μ
_{k}
. Find the value of the friction coefficient that will prevent mass m to slip and the value and the direction of the acceleration of mass m, suppose that mass m is much less than the mass of the cart.
From the free body diagram of the cart (mass M) we see that the force M g sinθ is accelerating the cart allong the slope. if this acceleration is marked as a and his direction is along the slope then:
M g sinθ = M a
⟶
a = g sinθ
As a consequence of the acceleration of the cart a force of
m a sinθ
is acting vertically upward on mass m , and the normal force on m (from the lower free body diagram) will be:
N = m
(
g − a sinθ
)
= m g cos
^{2}
θ
Mass m is attached to the cart only by the friction force f, once the acelleration force m a
_{m}
> f mass m will accelerate to the right.
From the free body diagram the equilibrium equations of m are:
a m cosθ − f = m a
_{m}
(1)
a
_{m}
= g sinθ cosθ − μ
_{k}
(g − a sinθ)
a
_{m}
= g cosθ (sinθ − μ
_{k}
cosθ)
(2)
In order that mass m will not slip we have a
_{m}
= 0 and μ is:
(3)
If we substitute the value a = g sinθ into eq. (3) we get:
μ ≥ tanθ
Q.
A small mass is located at the middle of a cart horizontal platform whose length is 1.2 meter, the cart is located on a tilted surface of sinθ = 0.6, the friction coefficient is μ
_{s}
= μ
_{k}
= 0.4 . Find the minimum value of friction coefficient that mass m will be at rest relative to the cart and the time needed to reach the end of the surface.
S.
Minimum friction coefficient is according to eq. (3)
μ = tanθ = tan (sin
^{-1}
0.6) = 0.75
The acceleration of mass m is from eq. (2):
a
_{m}
= 9.8 cos36.86 (0.6 − 0.4 * cos36.86) = tan (sin
^{-1}
0.6) = 2.24 m/s
^{2}